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Math Help - Partial Fraction and integrate

  1. #1
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    Partial Fraction and integrate

    ok it says to integrate. ....

    Integral (5x^4 - 2x^3 +2x^2 +4) / (x^3 - x^2) dx

    so since the numerator's degree is higher we have to do long divison right? i got 5x + 3 + (x^2+4)/(x^3-x^2)

    but now... what do we use to integrate? is it just the fraction portion of the answer to long division? im stuck...
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  2. #2
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    Re:

    Re:
    Attached Thumbnails Attached Thumbnails Partial Fraction and integrate-72.gif  
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  3. #3
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    thanks but how did you get a 9?
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  4. #4
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    Quote Originally Posted by runner07 View Post
    ok it says to integrate. ....

    Integral (5x^4 - 2x^3 +2x^2 +4) / (x^3 - x^2) dx

    so since the numerator's degree is higher we have to do long divison right? i got 5x + 3 + (x^2+4)/(x^3-x^2)

    but now... what do we use to integrate? is it just the fraction portion of the answer to long division? im stuck...
    In your long division you should have got
    5x +3 +[(5x^2 +4) /(x^3 -x^2)

    You forgot the coefficient 5 of the x^2 term.

    Now you need to decompose the rational or fraction portion of the integrand, because the degree of the numerator now is lower than that of the degree of the denominator. You cannot use long division anymore.

    Your title for your question is partial fraction, remember?

    (5x^2 +4) /(x^3 -x^2)
    (5x^2 +4) / (x^2)(x -1)

    (5x^2 +4) / (x^2)(x-1) = A/x +B/(x^2) +C/(x-1)
    Multiply both sides by (x^2)(x-1),
    5x^2 +4 = A(x)(x-1) +B(x-1) +C(x^2) -------------(i)

    When x = 0, in (i),
    0 +4 = 0 +B(-1) +0
    So, B = -4 ---------------------**

    When x = 1, in (i),
    5(1^2) +4 = 0 +0 +C(1^2)
    So, C = 9 -------------------------**

    When x = 2, and B = -4, and C = 9, in (i),
    5(2^2) +4 = A(2)(2-1) +(-4)(2-1) +9(2^2)
    24 = 2A -4 +36
    24 +4 -36 = 2A
    A = -8/2 = -4 -------------------**

    Therefore,
    INT.[(5x^4 - 2x^3 +2x^2 +4) / (x^3 - x^2)]dx

    = INT.[5x +3 -4/x -4/(x^2) +9/(x-1)]dx

    And so on, and so forth.....
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